Boosted67, yes in theory that is correct and it doesn't matter which side of the solenoid you connect the resistor.
Anna Conda is also correct although the use of methods other than a simple resistor depends on the circumstances. A resistor will always work but as a product of dropping the current it also produces heat. The power lost in the 94.5 Ohm resistor will be (I^2 * R) = .05*.05 * 94.5 = 0.24W which is quite small and even a physically small resistor would suffice. Be careful if you use a potentiometer/variable resistor though, if you do, add a fixed resistor in series with it, for example 75 Ohm fixed and 50 Ohm variable. It will still give you adjustment around the value you want but the heat will be shared between the fixed and variable resistors. If you use a pot alone, the heat is only dissipated in the part of the track carrying the current which might be short enough for it to overheat.
For larger currents a resistor will still work but the circuit can be made more efficient by using switching methods to limit it.
Please check the solenoid ratings again, I'm not saying they are wrong but it would be more usual for one with 5.5 Ohm resistance to operate at a higher current than 50mA.
Brian.